Physical Chemistry Third Edition

11.2 Forward Reactions with One Reactant 497

The Method of Initial Rates

In this method we compare data directly with the rate law instead of with an integrated

rate law. The reaction is followed for a short time∆tduring which there is a change

∆[A]in the concentration of a reactant, A:

∆[A][A]∆t−[A] 0 (11.2-22)

The time∆tmust be short enough that∆[A]is much smaller than[A]. The method

has the advantage that the reverse reaction is almost certain to be negligible. The initial

rate is approximated by a quotient of finite differences:

rinitial−

1

a

d[A]

dt

≈−

1

a

∆[A]

∆t

(11.2-23)

whereais the stoichiometric coefficient of the reactant A. If there is only one reactant

and if the rate law corresponds to orderα, the initial rate corresponds to Eq. (11.1-8):

rinitial−

1

a

d[A]

dt

kf[A]α 0 (11.2-24)

where [A] 0 is the initial concentration of the reactant A. We take the natural logarithm

of both sides of this equation

ln(rinitial)ln(kf)+αln([A] 0 ) (11.2-25)

To determine the order and the rate constant, one carries out several experiments

at the same temperature but with different values of[A] 0. A plot of the logarithm of

the initial rate as a function of the logarithm of the initial concentration is constructed

and a linear least-squares fit is performed. The slope of the line fitting the data points

is the order of the reaction, and the intercept is the logarithm of the rate constant.

A modification to the foregoing method would be to run the experiment over a longer

period of time and then to determine the value of∆[A]/∆tand the value of [A] at

different times and to use these values in a fit to Eq. (11.2-25).

25

29

28

27

26

24

In(initial concentration)

In(initial rate)

23 22

Figure 11.4 The Linear Plot for Exa-

mple 11.7.

EXAMPLE11.7

Following are data for the decomposition of ethyl chloride, C 2 H 5 Cl, at 500◦C. Find the order

of the reaction and the rate constant at this temperature.

Initial concentration /mol L−^1 Initial rate /mol L−^1 hour−^1

0.0500 0.00130

0.0400 0.00104

0.0300 0.00080

0.0200 0.00052

0.0100 0.00026

Solution

A linear least-squares fit of ln(rinitial) against ln([C 2 H 5 Cl] 0 ) gives

α 1 .00, ln(kf)− 3 .64, kf 0 .026 hour−^1

Figure 11.4 shows the graph of the data points and the least-squares line. The correlation

coefficient squared was equal to 1.000.